# Bond Enthalpy

> AP Chemistry · CED Unit 6 Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u6-bond-enthalpy/

This subtopic covers the definition of average bond enthalpy, calculation of reaction enthalpy via the bond enthalpy method, sign conventions, bond strength trends, and prediction of reaction thermicity for AP Chemistry Unit 6.

**Prerequisites:** Enthalpy change (ΔH) definition and sign conventions for endo/exothermic reactions; Balancing chemical equations for gas-phase reactions; Difference between average and exact molecular properties

## Learning objectives

- Define average bond enthalpy and recall sign conventions for bond breaking and formation
- Calculate reaction enthalpy from given average bond enthalpy values
- Relate bond enthalpy to bond order, bond length, and bond strength
- Justify whether a reaction is endothermic or exothermic using bond enthalpy arguments

## What Is Bond Enthalpy?

Bond enthalpy (also called bond energy or average bond dissociation enthalpy) is defined as the enthalpy change required to homolytically break one mole of a specific covalent bond in the gaseous state, averaged across many different compounds containing that bond. Common notation is $E(\text{X-Y})$ or $\text{bond enthalpy}$, with standard units of kilojoules per mole ($\text{kJ mol}^{-1}$).

Because bond enthalpies are averaged across different chemical environments, they are approximate values, not exact for any single molecule. This topic is a core skill in AP Chemistry Unit 6, appearing regularly in both multiple-choice and free-response sections.

**Average Bond Enthalpy** — Enthalpy change required to break one mole of a given covalent bond in the gaseous state, averaged across all compounds containing that bond.

*Notation:* $E(\text{X-Y})$, $\text{bond enthalpy}$

*Example:* $E(\text{C-H}) = 414 \text{ kJ mol}^{-1}$ is the average value across all organic molecules

> **tip**
>
> A core foundational rule: breaking any covalent bond is always endothermic (requires energy input, so ΔH for bond breaking is positive), while forming any covalent bond is always exothermic (releases energy, so ΔH for bond formation is negative). This rule underpins all bond enthalpy calculations.

## Calculating Reaction Enthalpy from Bond Enthalpies

To calculate the net enthalpy change of a reaction, we first account for all energy required to break all reactant bonds into gaseous atoms, then subtract the energy released when those atoms form all product bonds. Breaking bonds contributes positive energy, while forming bonds contributes negative energy, leading to the core formula:

$$\Delta H_{\text{rxn}} = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$$

This method only works for reactions where all reactants and products are in the gaseous state. Bond enthalpies do not account for extra enthalpy changes from phase changes of solids or liquids, so those reactions require Hess’s law or enthalpy of formation instead. Accurate counting of bonds is the most critical step for getting the correct result; drawing full Lewis structures for all species avoids miscounting.

**Worked example:** Calculate the enthalpy change for the complete combustion of one mole of gaseous methane, using the following bond enthalpy values: $E(\text{C-H}) = 414 \text{ kJ mol}^{-1}$, $E(\text{O=O}) = 498 \text{ kJ mol}^{-1}$, $E(\text{C=O}) = 799 \text{ kJ mol}^{-1}$, $E(\text{O-H}) = 463 \text{ kJ mol}^{-1}$.

1. Write the balanced gas-phase reaction:

   $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$$
2. Count and sum bonds broken in reactants: 4 C-H bonds (1 mol CH₄) + 2 O=O bonds (2 mol O₂):

   $$\sum E(\text{broken}) = (4 \times 414) + (2 \times 498) = 1656 + 996 = 2652 \text{ kJ}$$
3. Count and sum bonds formed in products: 2 C=O bonds (1 mol CO₂) + 4 O-H bonds (2 mol H₂O):

   $$\sum E(\text{formed}) = (2 \times 799) + (4 \times 463) = 1598 + 1852 = 3450 \text{ kJ}$$
4. Apply the formula to get the final enthalpy change:

   $$\Delta H_{\text{rxn}} = 2652 - 3450 = -798 \text{ kJ mol}^{-1}$$

> **tip**
>
> Always draw full Lewis structures before counting bonds — the most common error is miscounting the number of double bonds (e.g., two C=O bonds in CO₂, not one), which leads to an incorrect result.

## Bond Enthalpy, Strength, and Bond Length Trends

Bond enthalpy directly measures bond strength: the higher the bond enthalpy, the more energy required to break the bond, so the stronger the bond. For bonds between the same pair of atoms, bond enthalpy is inversely related to bond length: shorter bonds have higher bond enthalpy and are stronger. This trend follows bond order: as bond order increases (single → double → triple), bond length decreases and bond enthalpy increases. AP questions regularly ask to rank bonds by strength, or justify a given trend using bond enthalpy arguments, often in FRQ sections.

**Worked example:** Rank the following carbon-oxygen bonds from strongest to weakest, then explain the relationship between bond order, bond length, and bond enthalpy for this series: C-O in methanol ($E=358 \text{ kJ mol}^{-1}$), C=O in formaldehyde ($E=745 \text{ kJ mol}^{-1}$), C≡O in carbon monoxide ($E=1072 \text{ kJ mol}^{-1}$).

1. Recall that higher bond enthalpy equals stronger bond, so order by bond enthalpy from highest to lowest: C≡O (1072 kJ mol⁻¹) > C=O (745 kJ mol⁻¹) > C-O (358 kJ mol⁻¹).
2. Identify the bond order for each bond: C≡O has bond order 3, C=O has bond order 2, C-O has bond order 1.
3. For bonds between the same two atoms (carbon and oxygen here), as bond order increases, bond length decreases and bond enthalpy increases, resulting in stronger bonds.

> **tip**
>
> The inverse bond length-bond enthalpy trend only applies to bonds between the same pair of elements. You cannot use this trend to compare, for example, H-F and I-I, because the bonded atoms have different sizes and electronegativities.

## Predicting Reaction Thermicity from Bond Enthalpy

AP FRQ frequently asks to justify whether a reaction is endothermic or exothermic using bond enthalpy arguments, without requiring a full numerical calculation. If the total energy required to break reactant bonds is greater than the total energy released when forming product bonds, ΔH is positive (endothermic). If the total energy released from forming product bonds exceeds the energy required to break reactant bonds, ΔH is negative (exothermic). This skill is often tested in contextual questions about polymerization, fuel combustion, or bond cleavage reactions.

**Worked example:** The hydrogenation of ethene follows the reaction $\text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$. Given that the average bond enthalpy of a C-C single bond is greater than half the bond enthalpy of a C=C double bond, predict if the reaction is endothermic or exothermic and justify your answer.

1. Count bonds broken and formed: Bonds broken = 1 C=C (ethene) + 1 H-H (H₂); Bonds formed = 1 C-C (ethane) + 2 new C-H bonds.
2. Write the ΔH expression:

   $$\Delta H = [E(\text{C=C}) + E(\text{H-H})] - [E(\text{C-C}) + 2E(\text{C-H})]$$
3. Apply the given condition: $E(\text{C-C}) > \frac{E(\text{C=C})}{2}$, so $E(\text{C=C}) < 2E(\text{C-C})$. The total energy of bonds broken is less than the total energy of bonds formed.
4. ΔH = (smaller value) - (larger value) = negative, so the reaction is exothermic.

> **tip**
>
> When justifying the sign of ΔH on FRQ, you must explicitly compare total energy of bonds broken vs. bonds formed. You will not earn full credit for only stating that "bond breaking is endothermic and bond forming is exothermic" without the comparison.

## AP-Style Practice Problems

**Check your understanding**

Test your calculation skills with this multiple-choice question:

1. Which of the following correctly gives the enthalpy change for the reaction below, using the provided bond enthalpies? $$\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$$ Bond enthalpies: $E(\text{H-H}) = 436 \text{ kJ mol}^{-1}$, $E(\text{Cl-Cl}) = 242 \text{ kJ mol}^{-1}$, $E(\text{H-Cl}) = 431 \text{ kJ mol}^{-1}$.

   - -184 kJ mol⁻¹
   - +184 kJ mol⁻¹
   - -248 kJ mol⁻¹
   - +248 kJ mol⁻¹

   *Answer:* -184 kJ mol⁻¹

   *Why:* Correct: Sum of bonds broken = 436 + 242 = 678 kJ, sum of bonds formed = 2×431 = 862 kJ, ΔH = 678 - 862 = -184 kJ mol⁻¹. Reversing the formula gives the incorrect positive value.

**Worked example:** Ethene undergoes polymerization to form polyethylene according to the reaction: $n\text{ C}_2\text{H}_4(g) \rightarrow -(\text{CH}_2\text{CH}_2)_n - (g)$ (end groups are negligible for large $n$). Use the bond enthalpies $E(\text{C=C}) = 614 \text{ kJ mol}^{-1}$, $E(\text{C-C}) = 348 \text{ kJ mol}^{-1}$ to answer: (a) Calculate ΔH per mole of ethene monomer. (b) Justify the sign of ΔH. (c) If starting with liquid ethene, is the magnitude of ΔH larger or smaller? Explain.

1. (a) All C-H bonds are unchanged, so they do not contribute to ΔH. Per mole of monomer, 1 C=C bond is broken, and 1 full mole of new C-C bonds is formed:

   $$\Delta H = 614 - (2 \times 348) = -82 \text{ kJ mol}^{-1}$$
2. (b) ΔH is negative because the energy required to break the C=C bond (614 kJ/mol) is less than the energy released forming the two new C-C bonds (696 kJ/mol). Net energy is released, so the reaction is exothermic.
3. (c) The magnitude of ΔH will be smaller. Converting liquid ethene to gaseous ethene is endothermic, so the positive energy input partially offsets the negative polymerization enthalpy, leading to a smaller net magnitude.

**Worked example:** The Haber process produces ammonia for fertilizer: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. Bond enthalpies: $E(\text{N≡N}) = 945 \text{ kJ mol}^{-1}$, $E(\text{H-H}) = 436 \text{ kJ mol}^{-1}$, $E(\text{N-H}) = 391 \text{ kJ mol}^{-1}$. A plant claims the reaction releases usable energy. Calculate ΔH per mole of ammonia produced, and confirm if the claim is supported.

1. Calculate total bonds broken for the full reaction:

   $$\sum E(\text{broken}) = 945 + (3 \times 436) = 2253 \text{ kJ}$$
2. Calculate total bonds formed for 2 moles of NH₃:

   $$\sum E(\text{formed}) = 6 \times 391 = 2346 \text{ kJ}$$
3. Scale ΔH to per mole of ammonia:

   $$\Delta H_{\text{rxn}} = 2253 - 2346 = -93 \text{ kJ} \implies \Delta H_{\text{per mole}} = \frac{-93}{2} = -46.5 \text{ kJ mol}^{-1}$$
4. The negative ΔH confirms the reaction is exothermic and releases energy, so the plant's claim is supported.

## Common pitfalls

- **Wrong:** Counting one C=O bond in CO₂ instead of two
  - Why it fails: Students miscount multiple bonds within a single molecule
  - Correct: Always draw full Lewis structures for all species before counting bonds
- **Wrong:** Reversing the formula to ΔH = Σ E(formed) - Σ E(broken)
  - Why it fails: Confusion over the sign convention for bond formation
  - Correct: Use the mnemonic 'Break in, Make out': ΔH = energy in (breaking) minus energy out (making)
- **Wrong:** Using bond enthalpy for reactions with solid/liquid reactants/products
  - Why it fails: Forgetting bond enthalpies only apply to gas-phase bonds and ignore phase change enthalpy
  - Correct: Only use bond enthalpy method if all species are gaseous; use Hess's law/enthalpy of formation otherwise
- **Wrong:** Claiming shorter bonds are always stronger than longer bonds regardless of atoms
  - Why it fails: Overgeneralizing the inverse bond length-bond enthalpy trend
  - Correct: Only use the trend for bonds between the same two elements; compare given bond enthalpies directly for different atoms
- **Wrong:** Forgetting to multiply bond counts by stoichiometric coefficients
  - Why it fails: Counting bonds per molecule but not scaling to the reaction as written
  - Correct: Multiply bonds per molecule by the molecule's stoichiometric coefficient to get total moles of bonds
- **Wrong:** Treating bond enthalpy values as exact for a specific molecule
  - Why it fails: Forgetting bond enthalpies are averaged across many different compounds
  - Correct: Always note that ΔH calculated from bond enthalpy is an approximation for FRQ justifications

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Reaction Enthalpy | $\Delta H_{\text{rxn}} = \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$ | Only for all-gas-phase reactions; result is approximate |
| ΔH for Bond Breaking | $\Delta H = +E(\text{bond})$ | Always endothermic, positive sign |
| ΔH for Bond Forming | $\Delta H = -E(\text{bond})$ | Always exothermic, negative sign |
| Bond Enthalpy vs Strength | Higher $E(\text{bond})$ = stronger bond | Applies to all bonds |
| Bond Enthalpy vs Length | Higher $E(\text{bond})$ = shorter bond length | Only valid for bonds between the same two elements |
| Bond Enthalpy vs Bond Order | Higher bond order = higher $E(\text{bond})$ | Only for bonds between the same two elements |
| ΔH per mole of species | $\Delta H_{\text{per mole}} = \frac{\Delta H_{\text{rxn}}}{\text{stoichiometric coefficient}}$ | Used when asked for enthalpy per specific species |

## What's next

Bond enthalpy is a foundational skill for understanding reaction energy profiles, activation energy, and how covalent bond structure impacts reaction kinetics, which you will explore further in AP Chemistry Unit 5 (Kinetics) and Unit 6 (Thermodynamics). Without mastering bond enthalpy calculations and sign conventions, you will not be able to correctly calculate activation energy from bond energies or justify why some reactions have higher energy barriers than others. Beyond kinetics and thermodynamics, bond enthalpy is used to compare the stability of different allotropes (e.g., diamond vs graphite) and calculate the energy content of different fuels, a common context for AP FRQ questions. Building on this topic, you will next apply enthalpy calculation rules to other methods that work for non-gas phase reactions.

- [Enthalpy of Formation](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-of-formation/)
- [Hess's Law](https://www.owlsprep.com/study/ap-chemistry-u6-hess-s-law/)
- [Equilibrium Overview](https://www.owlsprep.com/study/ap-chemistry-u7-overview/)

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